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- Question
What will be the output of the program?
String a = "ABCD"; String b = a.toLowerCase(); b.replace('a','d'); b.replace('b','c'); System.out.println(b);
Options- A. abcd
- B. ABCD
- C. dccd
- D. dcba
- Correct Answer
- abcd
ExplanationString objects are immutable, they cannot be changed, in this case we are talking about the replace method which returns a new String object resulting from replacing all occurrences of oldChar in this string with newChar.b.replace(char oldChar, char newChar);
But since this is only a temporary String it must either be put to use straight away i.e.
System.out.println(b.replace('a','d'));
Or a new variable must be assigned its value i.e.
String c = b.replace('a','d');
More questions
- 1. What will be the output of the program?
System.out.println(Math.sqrt(-4D));
Options- A. -2
- B. NaN
- C. Compile Error
- D. Runtime Exception Discuss
Correct Answer: NaN
Explanation:
It is not possible in regular mathematics to get a value for the square-root of a negative number therefore a NaN will be returned because the code is valid.- 2. What will be the output of the program?
package foo; import java.util.Vector; /* Line 2 */ private class MyVector extends Vector { int i = 1; /* Line 5 */ public MyVector() { i = 2; } } public class MyNewVector extends MyVector { public MyNewVector () { i = 4; /* Line 15 */ } public static void main (String args []) { MyVector v = new MyNewVector(); /* Line 19 */ } }
Options- A. Compilation will succeed.
- B. Compilation will fail at line 3.
- C. Compilation will fail at line 5.
- D. Compilation will fail at line 15. Discuss
Correct Answer: Compilation will fail at line 3.
Explanation:
Option B is correct. The compiler complains with the error "modifier private not allowed here". The class is created private and is being used by another class on line 19.- 3. What will be the output of the program?
TreeSet map = new TreeSet(); map.add("one"); map.add("two"); map.add("three"); map.add("four"); map.add("one"); Iterator it = map.iterator(); while (it.hasNext() ) { System.out.print( it.next() + " " ); }
Options- A. one two three four
- B. four three two one
- C. four one three two
- D. one two three four one Discuss
Correct Answer: four one three two
Explanation:
TreeSet assures no duplicate entries; also, when it is accessed it will return elements in natural order, which typically means alphabetical.- 4. Which three statements are true?
- Assertion checking is typically enabled when a program is deployed.
- It is never appropriate to write code to handle failure of an assert statement.
- Assertion checking is typically enabled during program development and testing.
- Assertion checking can be selectively enabled or disabled on a per-package basis, but not on a per-class basis.
- Assertion checking can be selectively enabled or disabled on both a per-package basis and a per-class basis.
Options- A. 1, 2 and 4
- B. 2, 3 and 5
- C. 3, 4 and 5
- D. 1, 2 and 5 Discuss
Correct Answer: 2, 3 and 5
Explanation:
(1) is wrong. It's just not true.(2) is correct. You're never supposed to handle an assertion failure.
(3) is correct. Assertions let you test your assumptions during development, but the assertion code?in effect?evaporates when the program is deployed, leaving behind no overhead or debugging code to track down and remove.
(4) is wrong. See the explanation for (5) below.
(5) is correct. Assertion checking can be selectively enabled or disabled on a per-package basis. Note that the package default assertion status determines the assertion status for classes initialized in the future that belong to the named package or any of its "subpackages".
The assertion status can be set for a named top-level class and any nested classes contained therein. This setting takes precedence over the class loader's default assertion status, and over any applicable per-package default. If the named class is not a top-level class, the change of status will have no effect on the actual assertion status of any class.
- 5. What will be the output of the program?
class s implements Runnable { int x, y; public void run() { for(int i = 0; i < 1000; i++) synchronized(this) { x = 12; y = 12; } System.out.print(x + " " + y + " "); } public static void main(String args[]) { s run = new s(); Thread t1 = new Thread(run); Thread t2 = new Thread(run); t1.start(); t2.start(); } }
Options- A. DeadLock
- B. It print 12 12 12 12
- C. Compilation Error
- D. Cannot determine output. Discuss
Correct Answer: It print 12 12 12 12
Explanation:
The program will execute without any problems and print 12 12 12 12.- 6. What will be the output of the program?
public class ThreadTest extends Thread { public void run() { System.out.println("In run"); yield(); System.out.println("Leaving run"); } public static void main(String []argv) { (new ThreadTest()).start(); } }
Options- A. The code fails to compile in the main() method
- B. The code fails to compile in the run() method
- C. Only the text "In run" will be displayed
- D. The text "In run" followed by "Leaving run" will be displayed Discuss
Correct Answer: The text "In run" followed by "Leaving run" will be displayed
- 7. Which statement is true?
Options- A. A static method cannot be synchronized.
- B. If a class has synchronized code, multiple threads can still access the nonsynchronized code.
- C. Variables can be protected from concurrent access problems by marking them with the synchronized keyword.
- D. When a thread sleeps, it releases its locks. Discuss
Correct Answer: If a class has synchronized code, multiple threads can still access the nonsynchronized code.
Explanation:
B is correct because multiple threads are allowed to enter nonsynchronized code, even within a class that has some synchronized methods.A is incorrect because static methods can be synchronized; they synchronize on the lock on the instance of class java.lang.Class that represents the class type.
C is incorrect because only methods?not variables?can be marked synchronized.
D is incorrect because a sleeping thread still maintains its locks.
- 8. Which of the following are legal lines of code?
- int w = (int)888.8;
- byte x = (byte)1000L;
- long y = (byte)100;
- byte z = (byte)100L;
Options- A. 1 and 2
- B. 2 and 3
- C. 3 and 4
- D. All statements are correct. Discuss
Correct Answer: All statements are correct.
Explanation:
Statements (1), (2), (3), and (4) are correct. (1) is correct because when a floating-point number (a double in this case) is cast to an int, it simply loses the digits after the decimal.(2) and (4) are correct because a long can be cast into a byte. If the long is over 127, it loses its most significant (leftmost) bits.
(3) actually works, even though a cast is not necessary, because a long can store a byte.
- 9. What will be the output of the program?
int I = 0; label: if (I < 2) { System.out.print("I is " + I); I++; continue label; }
Options- A. I is 0
- B. I is 0 I is 1
- C. Compilation fails.
- D. None of the above Discuss
Correct Answer: Compilation fails.
Explanation:
The code will not compile because a continue statement can only occur in a looping construct. If this syntax were legal, the combination of the continue and the if statements would create a kludgey kind of loop, but the compiler will force you to write cleaner code than this.- 10. What will be the output of the program?
interface Foo141 { int k = 0; /* Line 3 */ } public class Test141 implements Foo141 { public static void main(String args[]) { int i; Test141 test141 = new Test141(); i = test141.k; /* Line 11 */ i = Test141.k; i = Foo141.k; } }
Options- A. Compilation fails.
- B. Compiles and runs ok.
- C. Compiles but throws an Exception at runtime.
- D. Compiles but throws a RuntimeException at runtime. Discuss
Correct Answer: Compiles and runs ok.
Explanation:
The variable k on line 3 is an interface constant, it is implicitly public, static, and final. Static variables can be referenced in two ways:Via a reference to any instance of the class (line 11)
Via the class name (line 12).
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- 1. What will be the output of the program?